# Why does $\langle a_-\alpha|\alpha\rangle = \alpha$ for harmonic oscillator?

Given $|\alpha\rangle$ a coherent state, why does $\langle a_-\alpha|\alpha\rangle = \alpha$? Doesn't the ladder operator lower the $\alpha$ so that it becomes $\sqrt{\alpha}\,\delta_{\alpha, \alpha_{-1}}$?

Maybe it's because that $\alpha$ can be imaginary? In that case how would you prove the above relation?

Although you don't say so in your question, it's clear from the context that $|\alpha\rangle$ is here being used to denote a so-called coherent state defined by the condition that it is an eigenvector of the lowering operator; \begin{align} a|\alpha\rangle = \alpha|\alpha\rangle. \end{align} It follows that \begin{align} \langle \alpha|a|\alpha\rangle = \alpha \langle \alpha|\alpha\rangle = \alpha \end{align} where in the last step we have assumed that the coherent state $|\alpha\rangle$ is normalized to one; \begin{align} \langle \alpha|\alpha\rangle = 1. \end{align}